Math 181/281. Rumbos Spring 2011 Page 1. Solutions to Assignment #5. Hence, the dynamical system, θ(t, p, q), for (t, p, q) R 3 corresponding = F.

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1 Math 181/281. Rumbos Spring 2011 Page 1 Solutions to Assignment #5 1. For real numbers a and b with a 2 + b 2 = 0, let F : R 2 R 2 be given b ( ) ( ) ( ) x ax b x F =, for all R 2. (1) bx + a (a) Explain wh the dnamical sstem, θ(t, p, q), for (t, p, q) R 3 corresponding to the field F exists. Answer: Note that the field F : R 2 R 2 defined in (1) is linear; hence, b the existence and uniqueness theorem developed in the lecture notes the IVP d ( ) x = F ( ) x(0) = (0) ( ) x ; ( ) p, q ( ) p has a solution for each R 2, which exists for all t R. q Hence, the dnamical sstem, θ(t, p, q), for (t, p, q) R 3 corresponding to the field F exists. (b) Prove that (0, 0) is the onl equilibrium point of the field F. Solution: Equilibrium points of the sstem in (2) are solutions to the equation ( ) ( ) x 0 F =, 0 or ( ) ( ) a b x = b a Now, since ( ) a b det = a 2 + b 2 = 0, b a (2) ( ) 0. (3) 0 the sstem in (3) has onl the trivial solution. Consequentl, (0, 0) is the onl equilibrium point of the sstem in (2).

2 Math 181/281. Rumbos Spring 2011 Page 2 (c) Define V (x, ) = x for all (x, ) R 2. Given (p, q) R 2 with (p, q) = (0, 0), define v(t) = V (θ(t, p, q)), for all t R; that is, the function v gives the values of V on the orbit γ (p,q). Compute v (t) and deduce from our result that if a < 0, then V decreases on γ (p,q) as t increases. What happens when a > 0. Solution: Write θ(t, p, q) = (x(t), (t)), for all t R. Then, and, b the Chain Rule, v(t) = V (x(t), (t)), for all t R, dv = V dx x + V d, where dx = ax b; We then have that dv d = bx + a. = 2x(ax b) + 2(bx + a) = 2ax 2 2bx + 2bx + 2a 2 = 2a(x ). We then have that dv = 2av. (4) It follows from (4) that, if a < 0 and (p, q) = (0, 0), then v (t) < 0 for all t R, so that V decreases on γ (p,q) as t increases. Similarl, if a > 0 and (p, q) = (0, 0), the V increases on γ (p,q) as t increases. (d) Compute the ω limit sets of γ (p,q), for (p, q) = (0, 0), in the cases a < 0 and a > 0.

3 Math 181/281. Rumbos Spring 2011 Page 3 Solution: obtain that Solving the differential equation in (4) for v(t) we v(t) = (p 2 + q 2 )e 2at, for all t R. (5) Noting that v(t) = θ(t, p, q) 2 for all t R, we obtain from (5) that θ(t, p, q) = (p, q) e at, for all t R. (6) Thus, if a < 0, it follows from (6) that lim θ(t m, p, q) = 0, m for an sequence, (t m ), of positive numbers for which t m as m. Thus, ω(γ (p,q) ) = {(0, 0)} for an (p, q) R 2, provided that a < 0. On the other hand, if a > 0, it follows from (6) that lim θ(t m, p, q) = +, m for an sequence, (t m ), of positive numbers for which t m as m, since (p, q) = 0. Consequentl, ω(γ (p,q) ) = for a > 0 and (p, q) = (0, 0). (e) Compute the α limit sets of γ (p,q), for (p, q) = (0, 0), in the cases a < 0 and a > 0. Solution: If a > 0, we obtain from (6) that lim θ(t m, p, q) = 0, m for an sequence, (t m ), of negative numbers for which t m as m. Thus, α(γ (p,q) ) = {(0, 0)} for an (p, q) R 2, for a > 0. On the other hand, if a < 0, then, using (6) again, we obtain that lim θ(t m, p, q) =, m for an sequence, (t m ), of negative numbers for which t m as m, since (p, q) = 0. Thus, α(γ (p,q) ) = for a < 0 and (p, q) = 0.

4 Math 181/281. Rumbos Spring 2011 Page 4 2. Assume that r = r(t) and θ = θ(t) are differentiable functions of t R, and define x(t) = r(t) cos θ(t) and (t) = r(t) sin θ(t) for all t R. Verif that = dx d cos θ + sin θ = 1 r d cos θ 1 dx sin θ. r Solution: Differentiate x = x(t) = r(t) cos θ(t) and = (t) = r(t) sin θ(t) with respect to t to obtain dx d = cos θ r sin θ ; = sin θ + r cos θ. The sstem in (8) can be written in matrix form as dx cos θ r sin θ d = sin θ r cos θ. (9) Since the 2 2 matrix in (9) is invertible, for r = 0, with inverse 1 cos θ r sin θ = 1 r cos θ r sin θ, r sin θ r cos θ sin θ cos θ we can solve the matrix equation in (9) for to obtain dx cos θ sin θ = 1 r sin θ 1 cos θ d, r which ields (7). (8) (7)

5 Math 181/281. Rumbos Spring 2011 Page 5 3. Use the transformation equations (7) derived in the previous problem to transform the sstem dx = ax b; (10) d = bx + a. into a sstem involving r and θ. Solution: Using the transformation equations in (7) we obtain the sstem = (ax b) cos θ + (bx + a) sin θ; = 1 r (bx + a) cos θ 1 (ax b) sin θ, r which is equivalent to = (ar cos θ br sin θ) cos θ + (br cos θ + ar sin θ) sin θ; or = (b cos θ + a sin θ) cos θ (a cos θ b sin θ) sin θ, = ar; = b. (11) (a) Solve the sstem for r and θ. Solution: Solving the differential equations in (11) for r and θ ields r(t) = c 1 e at, for t R, (12) and for some constants c 1 and c 2. θ(t) = bt + c 2, for t R, (13)

6 Math 181/281. Rumbos Spring 2011 Page 6 (b) Based on our formulas for r and θ, write down the general solution to the sstem (10) Solution: Substituting the expressions for r and θ in (12) and (13), respectivel, into x = r cos θ leads to or x(t) = c 1 e at cos(bt + c 2 ), x(t) = e at [c 1 cos c 2 cos(bt) c 1 sin c 2 sin(bt)], (14) where we have used the trigonometric identit cos(α + β) = cos α cos β sin α sin β. Similarl, from (12), (13) and = r sin θ we obtain that (t) = e at [c 1 cos c 2 sin(bt) + c 1 sin c 2 cos(bt)], (15) where we have used the trigonometric identit sin(α + β) = sin α cos β + cos α sin β. Renaming c 1 cos c 2 = p and c 2 sin c 2 = q, we obtain the general solution to the sstem in (10): { x(t) = e at [p cos(bt) q sin(bt)]; (t) = e at [p sin(bt) + q cos(bt)], or, in matrix form, ( ) ( x(t) cos(bt) sin(bt) = e at (t) sin(bt) cos(bt) ) ( p q ), for all t R. (16) (c) Use our result in the previous part to obtain the dnamical sstem, θ(t, p, q), for (t, p, q) R 3, for the sstem in (10). Explain wh this is the same sstem as the one mentioned in Part (a) of Problem 1. Solution: We obtain from (16) that ( ) ( ) cos(bt) sin(bt) p θ(t, p, q) = e at, for all t R, (17) sin(bt) cos(bt) q

7 Math 181/281. Rumbos Spring 2011 Page 7 since θ(0, p, q) = ( ) p. (18) q B the uniqueness of the solution to the sstem in (10) subject to the initial condition in (18), it follows that the dnamical sstem given in (17) is the same that was mentioned in Part (a) of Problem Assume that b > 0 and a = 0 in the two dimensional sstem (10). (a) Based on our solution to the previous problem in terms of r and θ, sketch a possible non trivial orbit of the sstem. Compute the α limit set of the orbit. What is the ω limit set of the orbit? Solution: Assume that b > 0 and a > 0, then the orbit of a point (p, q) R 2 is a spiral moving awa from the origin in a counterclockwise sense. Figure 1 shows a tpical orbit for the case a > 0 and b > 0. x Figure 1: Tpical Orbit of Sstem (10) for a > 0, b > 0. For this particular orbit the α limit set is (0, 0) and the ω limit set is empt. (b) Assume that a < 0 in the two dimensional sstem (10). Based on our solution in terms of r and θ resulting from the transformation equations (7), sketch a possible non trivial orbit of the sstem. Compute the ω limit set of the orbit. What is the α limit set of the orbit? Solution: Figure 2 on page 8 shows a tpical orbit of the sstem in (10) for the case a < 0 and b > 0. The ω limit set of this orbit is {(0, 0)}, while the α limit set is empt.

8 Math 181/281. Rumbos Spring 2011 Page 8 x Figure 2: Tpical Orbit of Sstem (10) for a < 0, b > Assume that b > 0 and a = 0 in the two dimensional sstem (10). Sketch the phase portrait of the sstem. What can ou sa about the nontrivial orbits? What do ou conclude about the solutions of the sstem? Solution: Figure 3 shows the phase portrait of the sstem in (10) for a = 0 and b > 0. The nontrivial orbits are concentric circles about the origin in R 2 oriented in the counterclockwise sense. Nontrivial solutions are periodic functions of period, T determined b '$ r &% x Figure 3: Phase Portrait of the sstem in (10) for a = 0, b > 0. bt = 2π, so that T = 2π b.